Download AMC 12A

Solutions Pamphlet
MAA American Mathematics Competitions
68th Annual
AMC 12A
American Mathematics Competition 12A
Tuesday, February 7, 2017
This Pamphlet gives at least one solution for each problem on this year’s competition
and shows that all problems can be solved without the use of a calculator. When more
than one solution is provided, this is done to illustrate a significant contrast in methods,
e.g., algebraic versus geometric, computational versus conceptual, elementary versus
advanced. These solutions are by no means the only ones possible, nor are they superior
to others the reader may devise.
We hope that teachers will inform their students about these solutions, both as
illustrations of the kinds of ingenuity needed to solve nonroutine problems and as
examples of good mathematical exposition. However, the publication, reproduction or
communication of the problems or solutions for this contest during the period when students
are eligible to participate seriously jeopardizes the integrity of the results. Dissemination
at any time via copier, telephone, email, internet, or media of any type is a violation of
the competition rules.
Correspondence about the problems/solutions for this AMC 12 and orders for any
publications should be addressed to:
MAA American Mathematics Competitions
Attn: Publications, PO Box 471, Annapolis Junction, MD 20701
Phone 800.527.3690 | Fax 240.396.5647 | [email protected]
The problems and solutions for this AMC 12 were prepared by
MAA’s Subcommittee on the AMC10/AMC12 Exams, under the direction of the
co-chairs Jerrold W. Grossman and Carl Yerger.
© 2017 Mathematical Association of America
2017 AMC 12A Solutions
2
1. Answer (D): The cheapest popsicles cost $3.00 ÷ 5 = $0.60 each.
Because 14 · $0.60 = $8.40 and Pablo has just $8, he could not pay for
14 popsicles even if he were allowed to buy partial boxes. The best
he can hope for is 13 popsicles, and he can achieve that by buying
two 5-popsicle boxes (for $6) and one 3-popsicle box (for $2).
OR
If Pablo buys two single popsicles for $1 each, he could have bought
a 3-popsicle box for the same amount of money. Similarly, if Pablo
buys three single popsicles or both one 3-popsicle box and one single
popsicle, he could have bought a 5-popsicle box for the same amount
of money. If Pablo buys two 3-popsicle boxes, he could have bought
a 5-popsicle box and a single popsicle for the same amount of money.
The previous statements imply that a maximum number of popsicles
for a given amount of money can be obtained by buying either at most
one single popsicle and the rest 5-popsicle boxes, or a single 3-popsicle
box and the rest 5-popsicle boxes. When Pablo has $8, he can obtain
the maximum number of popsicles by buying two 5-popsicle boxes and
one 3-popsicle box. This gives a total of 2 · 5 + 1 · 3 = 13 popsicles.
2. Answer (C): Let the two numbers be x and y. Then x + y = 4xy.
Dividing this equation by xy gives y1 + x1 = 4. One such pair of
numbers is x = 13 , y = 1.
3. Answer (B): The given statement is logically equivalent to its
contrapositive: If a student did not receive an A on the exam, then
the student did not get all the multiple choice questions right, which
means that he got at least one of them wrong. None of the other
statements follows logically from the given implication; the teacher
made no promises concerning students who did not get all the multiple
choice questions right. In particular, a statement does not imply its
inverse or its converse; and the negation of the statement that Lewis
got all the questions right is not the statement that he got all the
questions wrong.
4. Answer (A): If the square had side length x, then Jerry’s path had
length 2x, and Silvia’s √
path along the diagonal, by the Pythagorean
Theorem, had length 2x. Therefore Silvia’s trip was shorter by
2017 AMC 12A Solutions
√
2x − 2x, and the required percentage is
√
√
2
2x − 2x
=1−
≈ 1 − 0.707 = 0.293 = 29.3%.
2x
2
3
The closest of the answer choices is 30%.
5. Answer (B): Each of the 20 people who know each other shakes
hands with 10 people. Each of the 10 people who know no one shakes
hands with 29 people. Because each handshake involves two people,
the number of handshakes is 12 (20 · 10 + 10 · 29) = 245.
6. Answer (B): Four rods can form a quadrilateral with positive area
if and only if the length of the longest rod is less than the sum of the
lengths of the other three. Therefore if the fourth rod has length n cm,
then n must satisfy the inequalities 15 < 3 + 7 + n and n < 3 + 7 + 15,
that is, 5 < n < 25. Because n is an integer, it must be one of the 19
integers from 6 to 24, inclusive. However, the rods of lengths 7 cm
and 15 cm have already been chosen, so the number of rods that Joy
can choose is 19 − 2 = 17.
7. Answer (B): It is clear after listing the first few values, f (1) = 2,
f (2) = f (1) + 1 = 3, f (3) = f (1) + 2 = 4, f (4) = f (3) + 1 = 5,
and so on, that f (n) = n + 1 for all positive integers n. Indeed,
the function is uniquely determined by the recursive description, and
the function defined by f (n) = n + 1 fits the description. Therefore
f (2017) = 2018.
8. Answer (D): Let h = AB. The region consists of a solid circular
cylinder of radius 3 and height h, together with two solid hemispheres
of radius 3 centered at A and B. The volume of the cylinder is
π · 32 · h = 9πh, and the two hemispheres have a combined volume of
4
3
3 π · 3 = 36π. Therefore 9πh + 36π = 216π, and h = 20.
9. Answer (E): Suppose that the two larger quantities are the first
and the second. Then 3 = x + 2 ≥ y − 4. This is equivalent to x = 1
and y ≤ 7, and its graph is the downward-pointing ray with endpoint
(1, 7). Similarly, if the two larger quantities are the first and third,
then 3 = y − 4 ≥ x + 2. This is equivalent to y = 7 and x ≤ 1, and its
graph is the leftward-pointing ray with endpoint (1, 7). Finally, if the
2017 AMC 12A Solutions
4
two larger quantities are the second and third, then x + 2 = y − 4 ≥ 3.
This is equivalent to y = x + 6 and x ≥ 1, and its graph is the ray
with endpoint (1, 7) that points upward and to the right. Thus the
graph consists of three rays with common endpoint (1, 7).
13
10
7
4
1
−4
−1
1
4
7
10
Note: This problem is related to a relatively new area of mathematics
called tropical geometry.
10. Answer (C): Half of the time Laurent will pick a number between
2017 and 4034, in which case the probability that his number will
be greater than Chloé’s number is 1. The other half of the time, he
will pick a number between 0 and 2017, and by symmetry his number
will be the larger one in half of those cases. Therefore the requested
probability is 12 · 1 + 12 · 12 = 34 .
OR
The choices of numbers can be represented in the coordinate plane by
points in the rectangle with vertices at (0, 0), (2017, 0), (2017, 4034),
and (0, 4034). The portion of the rectangle representing the event that
Laurent’s number is greater than Chloé’s number is the portion above
the line segment with endpoints (0, 0) and (2017, 2017). This area is
3
3
4 of the area of the entire rectangle, so the requested probability is 4 .
11. Answer (D): If the polygon has n sides and the degree measure
of the forgotten angle is α, then (n − 2)180 = 2017 + α. Because
0 < α < 180,
2017 < (n − 2)180 < 2197,
2017 AMC 12A Solutions
5
which implies that n = 14, the angle sum is 2160, and α = 143. To
see that such a polygon exists, draw a circle and a central angle of
measure 143◦ , and divide the minor arc spanned by the angle into 12
small arcs. The polygon is then formed by the two radii and 12 small
chords, as illustrated.
12. Answer (B): Horse k will again be at the starting point after t
minutes if and only if k is a divisor of t. Let I(t) be the number of
integers k with 1 ≤ k ≤ 10 that divide t. Then I(1) = 1, I(2) = 2,
I(3) = 2, I(4) = 3, I(5) = 2, I(6) = 4, I(7) = 2, I(8) = 4, I(9) = 3,
I(10) = 4, I(11) = 1, and I(12) = 5. Thus T = 12 and the requested
sum of digits is 1 + 2 = 3.
13. Answer (B): Let d be the requested distance in miles, and suppose
that Sharon usually drives at speed r in miles per hour. Then dr = 3.
The total time in hours for Sharon’s trip with the snowstorm is then
1
2
23
d
3d
3d
r + r−20 = 5 . Because r = 3, this reduces to
1+
2
3
r
d
−
20
d
=1+
2
3
1
3
−
20
d
=
23
.
5
Solving for d gives d = 135.
OR
The last 23 of the drive takes 276 − 31 · 180 = 216 minutes, which is
216
2
60 hours. If r is the original speed in miles per hour, then 3 of the
216
distance is both 2r and 60 · (r − 20). Setting these expressions equal
and solving yields r = 45. Therefore the original speed is 45 miles
per hour, and the requested distance is 3 · 45 = 135 miles.
2017 AMC 12A Solutions
6
14. Answer (C): Let X be the set of ways to seat the five people in
which Alice sits next to Bob. Let Y be the set of ways to seat the
five people in which Alice sits next to Carla. Let Z be the set of ways
to seat the five people in which Derek sits next to Eric. The required
answer is 5! − |X ∪ Y ∪ Z|. The Inclusion–Exclusion Principle gives
|X∪Y ∪Z| = |X|+|Y |+|Z| − |X∩Y |+|X∩Z|+|Y ∩Z| +|X∩Y ∩Z|.
Viewing Alice and Bob as a unit in which either can sit on the other’s
left side shows that there are 2 · 4! = 48 elements of X. Similarly
there are 48 elements of Y and 48 elements of Z. Viewing Alice, Bob,
and Carla as a unit with Alice in the middle shows that |X ∩ Y | =
2 · 3! = 12. Viewing Alice and Bob as a unit and Derek and Eric as
a unit shows that |X ∩ Z| = 2 · 2 · 3! = 24. Similarly |Y ∩ Z| = 24.
Finally, there are 2 · 2 · 2! = 8 elements of X ∩ Y ∩ Z. Therefore
|X ∪ Y ∪ Z| = (48 + 48 + 48) − (12 + 24 + 24) + 8 = 92, and the answer
is 120 − 92 = 28.
OR
There are three cases based on where Alice is seated.
• If Alice takes the first or last chair, then Derek or Eric must be
seated next to her, Bob or Carla must then take the middle chair,
and either of the remaining two individuals can be seated in either of
the other two chairs. This gives a total of 24 = 16 arrangements.
• If Alice is seated in the second or fourth chair, then Derek and Eric
will take the seats on her two sides, and this can be done in two ways.
Bob and Carla can be seated in the two remaining chairs in two ways,
which yields a total of 23 = 8 arrangements.
• If Alice sits in the middle chair, then Derek and Eric will be seated
on her two sides, with Bob and Carla seated in the first and last
chairs. This results in 22 = 4 arrangements.
Thus there are 16 + 8 + 4 = 28 possible arrangements in total.
15. Answer (D): For 0 < x < π2 all three terms are positive, and f (x)
is undefined when x = π2 . For π2 < x < 3π
4 , the term 3 tan x is less
than −3 and dominates the other two terms, so f (x) < 0 there. For
3π
4 ≤ x < π, | cos(x)| ≥ | sin(x)| and cos x and tan x are negative, so
sin x + 2 cos x + 3 tan x < 0. Therefore there is no positive solution
of f (x) = 0 for x < π. Because the range
√ of f includes all values
3
between f (π) = −2 < 0 and f ( 5π
)
=
−
4
2 2 + 3 > −1.5 · 1.5 + 3 > 0
],
the
smallest
positive
solution of f (x) = 0 lies
on the interval [π, 5π
4
2017 AMC 12A Solutions
between π and
is (3, 4).
5π
4 .
7
Because π > 3 and
5π
4
< 4, the requested interval
16. Answer (B): Let C be the center of the largest semicircle, and let
r denote the radius of the circle centered at P . Note that P A = 2 + r,
P C = 3 − r, P B = 1 + r, AC = 1, BC = 2, and AB = 3. Let F
be the foot of the perpendicular from P to AB, let h = P F , and let
x = CF . The Pythagorean Theorem in P AF , P CF , and P BF
gives
h2 = (2 + r)2 − (1 + x)2 = (3 − r)2 − x2 = (1 + r)2 − (2 − x)2 .
This reduces to two linear equations in r and x, whose solution is
r = 76 , x = 97 .
r
2
J
A
P r
r
h
C
x
F
1
B
K
OR
With the notation and observations above, apply Heron’s Formula to
P CB and P AC, noting that the former has twice the area of the
latter and each has semiperimeter 3. Thus
3 · 1 · r · (r − 2) = 2 3 · 2 · r · (r − 1),
from which it follows that r = 67 .
OR
With the notation and observations above, let θ = ∠P AB. The Law
of Cosines applied to P AC gives
(3 − r)2 = (2 + r)2 + 12 − 2 · (2 + r) · 1 · cos θ,
2017 AMC 12A Solutions
8
and simplifying yields (2 + r) cos θ = 5r − 2. Applying the Law of
Cosines to P AB gives
(1 + r)2 = (2 + r)2 + 32 − 2 · (2 + r) · 3 · cos θ,
and simplifying yields 3(2 + r) cos θ = r + 6. Hence r + 6 = 3(5r − 2),
so r = 67 .
17. Answer (D): The complex numbers z such that z 24 = 1 are the
roots of z 24 − 1 = (z 6 − 1)(z 6 + 1)((z 6 )2 + 1). The factors can have
at most 6, 6, and 12, roots, respectively. Because z 24 − 1 has 24
distinct roots, the factors do actually have 6, 6, and 12 distinct roots,
respectively. The six roots of the first factor satisfy z 6 = 1, and the
six roots of the second factor satisfy z 6 = −1. The twelve roots of the
third factor satisfy (z 6 )2 = −1, so z 6 is never real in this case. There
are 6 + 6 = 12 roots such that z 6 is real.
OR
The complex values of z such that z 24 = 1 are the 24th roots of unity.
1
These values can be written in the form e 12 kπi , where k is an integer
between 0 and 23, inclusive. By Euler’s Theorem,
1
1
1
kπ + i sin
kπ .
z 6 = e 2 kπi = cos
2
2
This quantity is a real number if and only if sin 12 kπ = 0, which
occurs if and only if k is even. There are therefore 12 complex values
of z such that z 6 is real.
18. Answer (D): Note that S(n + 1) = S(n) + 1 unless the numeral
for n ends with a 9. Moreover, if the numeral for n ends with exactly
k 9s, then S(n + 1) = S(n) + 1 − 9k. Thus the possible values of
S(n + 1) when S(n) = 1274 are all of the form 1275 − 9k, where
k ∈ {0, 1, 2, 3, . . . , 141}. Of the choices, only 1239 can be formed in
this manner, and S(n + 1) will equal 1239 if, for example, n consists
of 4 consecutive 9s preceded by 1238 1s.
OR
The value of a positive integer is congruent to the sum of its digits
modulo 9. Therefore n ≡ S(n) = 1274 ≡ 5 (mod 9), so S(n + 1) ≡
n + 1 ≡ 6 (mod 9). Of the given choices, only 1239 meets this requirement.
2017 AMC 12A Solutions
9
x
19. Answer (D): In the first figure F EB ∼ DCE, so 3−x
= 4−x
x
and x = 12
7 . In the second figure, the small triangles are similar to
the large one, so the lengths of the portions of the side of length 3 are
x
12 37
37
as shown. Solving 35 y + 54 y = 3 yields y = 60
37 . Thus y = 7 · 60 = 35 .
C
4−x
D
E
x
x
A
y
F
3−x
y
B
3
5y
5
4y
20. Answer (E):
Let u = logb a. Because u2017 = 2017u, either u = 0
√
2016
2017. If √
u = 0, then a = 1 and b√can be any integer from
or u = ±
2016
2017
, where again b can
2 to 200. If u = ± 2016 2017, then a = b ±
be any integer from 2 to 200. Therefore there are 3 · 199 = 597 such
ordered pairs.
21. Answer (D): Because −1 is a root of 10x + 10, −1 is added to S.
Then 1 is also added to S, because it is a root of (−1)x10 + (−1)x9 +
· · · + (−1)x + 10. At this point −10, a root of 1 · x + 10, can be
added to S. Because 2 is a root of 1 · x3 + 0 · x2 + 1 · x + (−10),
and −2 is a root of 1 · x + 2, both 2 and −2 can be added to S.
The polynomials 2x + (−10) and 2x + 10 allow 5 and −5 into S. At
this point S = {0, ±1, ±2, ±5, ±10}. No more elements can be added
to S, because by the Rational Root Theorem, any integer root of a
polynomial with integer coefficients whose constant term is a factor
of 10 must be a factor of 10. Therefore S contains 9 elements.
Note: It is not true that in general if S starts with {0, c} then all
factors of c can be added to S. For example, applying the procedure
to {0, 35} gives only {0, ±1, ±35}, although of course it takes some
argument to rule out ±5 and ±7.
22. Answer (E):
Let A = {(1, 0), (0, 1), (−1, 0), (0, −1)}, let C =
{(0, 0)}, and let I = {(1, 1), (−1, 1), (−1, −1), (1, −1)}. A particle
2017 AMC 12A Solutions
10
in A will move to A with probability 28 , to C with probability 18 , to
I with probability 28 , and to an interior point of a side of the square
with probability 38 . Similarly, a particle in C will move to A with
probability 48 and to I with probability 48 ; and a particle in I will
move to A with probability 28 , to C with probability 18 , to a corner
of the square with probability 18 , and to an interior point of a side
of the square with probability 48 . Let a, c, and i be the probabilities
that the particle will first hit the square at a corner, given that it is
currently in A, C, and I, respectively. The transition probabilities
noted above lead to the following system of equations.
2
a+
8
4
c= a+
8
2
i= a+
8
a=
1
c+
8
4
i
8
1
c+
8
2
i
8
1
8
4
1
, c = 35
, and
This system can be solved by elimination to yield a = 14
11
i = 70 . The required fraction is c, whose numerator and denominator
sum to 39.
23. Answer (C):
Let q be the additional root of f (x). Then
f (x) = (x − q)(x3 + ax2 + x + 10)
= x4 + (a − q)x3 + (1 − qa)x2 + (10 − q)x − 10q.
Thus 100 = 10 − q, so q = −90 and c = −10q = 900. Also 1 = a − q =
a + 90, so a = −89.
It follows,
using the factored form of f shown
above, that f (1) = 1−(−90) ·(1−89+1+10) = 91·(−77) = −7007.
24. Answer (A): Because Y E and EF are parallel to AD and AC,
respectively, XEY ∼ XAD and XEF ∼ XAC. Therefore
XD
XY
=
XE
XA
It follows that
and
XF
XC
=
.
XE
XA
XC
XF
=
.
XD
XY
The Power of a Point Theorem applied to circle O and point X implies
that XC · XG = XD · XB. Together with the previous equation this
2017 AMC 12A Solutions
11
implies that XF · XG = XB · XY . Let d = BD; then DX = 14 d and
BY = 11
36 d. It follows that
XF · XG = XB · XY = (BD − DX) · (BD − DX − BY )
1
11
1
d− d− d
= d− d
4
4
36
3 4 2
d2
= · d = .
4 9
3
F
E
G
D
X
Y
A
B
C
To determine d, note that because ABCD is a cyclic quadrilateral it
follows that α = ∠BAD = π − ∠DCB. Applying the Law of Cosines
to ABD and BCD yields
cos α =
32 + 82 − d2
73 − d2
AB 2 + AD2 − BD2
=
=
,
2 · AB · AD
2·3·8
48
and
− cos α = cos(π−α) =
2 2 + 62 − d 2
40 − d2
CB 2 + CD2 − BD2
=
=
.
2 · CB · CD
2·2·6
24
2017 AMC 12A Solutions
Therefore
12
73 − d2
d2 − 40
=
,
48
24
and solving for d2 gives d2 = 51. Hence XF · XG = 13 d2 = 17.
√
√
25. Answer
√ (E): If zj is an element of the set A = { 2i, − 2i}, then
|zj | = 2. Otherwise zj is an element of
1
1
1
1
√
√
√
√
B =V \A=
(1 + i),
(−1 + i),
(1 − i),
(−1 − i)
8
8
8
8
12
and |zj | = 12 . It follows that |P | = j=1 |zj | = 1 exactly when 8 of
the 12 factors zj are in A and 4 of the factors are in B. The product
of 8 complex numbers each of which is in A is a real number, either
16 or −16. The product of 4 numbers each ofwhich is in B is one
12
1
1
1
1
of 16
, 16
i, − 16
, or − 16
i. Thus a product P = j=1 zj is −1 exactly
when 8 of the zj are from A, 4 of the zj are from B, and the last of
the 4 elements from B is chosen so that the product is −1 rather than
i, −i, or 1. Because the probability is 13 that a particular factor zj is
from A, the probability is 23 that a particular factor zj is from B, and
the probability is 16 that a particular factor zj is a specific element of
V , the probability that the product P will be −1 is given by
8 3 1
12
12 · 11 · 10 · 9 1 23 1
22 · 5 · 11
2
1
=
· 8· 3· =
.
4
3
3
6
4·3·2·1
3 3 6
310
Problems and solutions were contributed by Bernardo Abrego, Stephen
Adams, Paul Baird-Smith, Thomas Butts, Barb Currier, Steven Davis,
Marta Eso, Silvia Fernandez, Peter Gilchrist, Jerrold Grossman, Jonathan
Kane, Joe Kennedy, Michael Khoury, Matthew McMullen, Steven Miller,
Hugh Montgomery, Joachim Rebholz, Mark Saul, Roger Waggoner, Dave
Wells, and Carl Yerger.
The
MAA American Mathematics Competitions
are supported by
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Akamai Foundation
American Mathematical Society
American Statistical Association
Ansatz Capital
Army Educational Outreach Program
Art of Problem Solving
Casualty Actuarial Society
Conference Board of the Mathematical Sciences
The DE Shaw Group
Dropbox
Expii, Inc
IDEA MATH, LLC
Jane Street Capital
MathWorks
Mu Alpha Theta
National Council of Teachers of Mathematics
Simons Foundation
Society for Industrial and Applied Mathematics
Star League
Susquehanna International Group
Tudor Investment Corp
Two Sigma